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  • Codeforces Round #429 (Div. 2)

    A. Generous Kefa
    time limit per test
    2 seconds
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    One day Kefa found n baloons. For convenience, we denote color of i-th baloon as si — lowercase letter of the Latin alphabet. Also Kefa has k friends. Friend will be upset, If he get two baloons of the same color. Kefa want to give out all baloons to his friends. Help Kefa to find out, can he give out all his baloons, such that no one of his friens will be upset — print «YES», if he can, and «NO», otherwise. Note, that Kefa's friend will not upset, if he doesn't get baloons at all.

    Input

    The first line contains two integers n and k (1 ≤ n, k ≤ 100) — the number of baloons and friends.

    Next line contains string s — colors of baloons.

    Output

    Answer to the task — «YES» or «NO» in a single line.

    You can choose the case (lower or upper) for each letter arbitrary.

    Examples
    input
    4 2
    aabb
    output
    YES
    input
    6 3
    aacaab
    output
    NO
    Note

    In the first sample Kefa can give 1-st and 3-rd baloon to the first friend, and 2-nd and 4-th to the second.

    In the second sample Kefa needs to give to all his friends baloons of color a, but one baloon will stay, thats why answer is «NO».

     hash映射过去,有个字母大于k就是upset

    #include <bits/stdc++.h>
    using namespace std;
    int a[256];
    int main()
    {
        int n,k;
        cin>>n>>k;
        string s;
        cin>>s;
        for(int i=0; s[i]; i++)
            a[s[i]]++;
        for(int i=0; i<256; i++)
            if(a[i]>k)
            {
                return 0*printf("NO");
            }
        printf("YES");
        return 0;
    }
    B. Godsend
    time limit per test
    2 seconds
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    Leha somehow found an array consisting of n integers. Looking at it, he came up with a task. Two players play the game on the array. Players move one by one. The first player can choose for his move a subsegment of non-zero length with an odd sum of numbers and remove it from the array, after that the remaining parts are glued together into one array and the game continues. The second player can choose a subsegment of non-zero length with an even sum and remove it. Loses the one who can not make a move. Who will win if both play optimally?

    Input

    First line of input data contains single integer n (1 ≤ n ≤ 106) — length of the array.

    Next line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 109).

    Output

    Output answer in single line. "First", if first player wins, and "Second" otherwise (without quotes).

    Examples
    input
    4
    1 3 2 3
    output
    First
    input
    2
    2 2
    output
    Second
    Note

    In first sample first player remove whole array in one move and win.

    In second sample first player can't make a move and lose.

     去奇数偶数和的子段,这个就是根据奇偶分析啊,有奇数的第一个人就会赢

    听说数据量会卡cin

    #include <bits/stdc++.h>
    using namespace std;
    const int N=1e6+5;
    int a[N];
    int main()
    {
        int n,f=0;
        scanf("%d",&n);
        for(int i=1;i<=n;i++){
        scanf("%d",&a[i]);
        if(a[i]&1)f=1;
        }
        printf(f?"First":"Second");
        return 0;
    }
    C. Leha and Function
    time limit per test
    2 seconds
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    Leha like all kinds of strange things. Recently he liked the function F(n, k). Consider all possible k-element subsets of the set [1, 2, ..., n]. For subset find minimal element in it. F(n, k) — mathematical expectation of the minimal element among all k-element subsets.

    But only function does not interest him. He wants to do interesting things with it. Mom brought him two arrays Aand B, each consists of m integers. For all i, j such that 1 ≤ i, j ≤ m the condition Ai ≥ Bj holds. Help Leha rearrange the numbers in the array A so that the sum  is maximally possible, where A' is already rearranged array.

    Input

    First line of input data contains single integer m (1 ≤ m ≤ 2·105) — length of arrays A and B.

    Next line contains m integers a1, a2, ..., am (1 ≤ ai ≤ 109) — array A.

    Next line contains m integers b1, b2, ..., bm (1 ≤ bi ≤ 109) — array B.

    Output

    Output m integers a'1, a'2, ..., a'm — array A' which is permutation of the array A.

    Examples
    input
    5
    7 3 5 3 4
    2 1 3 2 3
    output
    4 7 3 5 3
    input
    7
    4 6 5 8 8 2 6
    2 1 2 2 1 1 2
    output
    2 6 4 5 8 8 6

     类似于基数排序,蜜汁wa了n发,好像是输出出了差错

    #include <bits/stdc++.h>
    using namespace std;
    const int N=2e5+5;
    struct SB
    {
        int a,b;
        bool friend operator < (SB x,SB y)
        {
            return x.a<y.a||x.a==y.a&&x.b<y.b;
        }
    } S[N];
    int A[N];
    int c[N];
    int main()
    {
        int n,f=0;
        scanf("%d",&n);
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&c[i]);
        }
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&S[i].a);
            S[i].b=i;
        }
        sort(c+1,c+n+1,greater<int>());
        sort(S+1,S+n+1);
        for(int i=1; i<=n; i++)
            A[S[i].b]=c[i];
        printf("%d",A[1]);
        for(int i=2; i<=n; i++)
            printf(" %d",A[i]);
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/BobHuang/p/7395486.html
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